Electrical – Forward Voltage drop of laser line module

I see a fundamental units problem in your math. You are using the symbol m which stands for milli (10^-3). The actual energies are µ (10^-6). You are off by a factor of 10^-3. Since most keyboards do not have a symbol for µ (micro), people often revert to an m (milli), and end up confusing the units. Hey, they both start with the same letter, right? :)

Additionally, there is the question of "wall plug efficiency". While a diode may 60mW, even military grade lasers only have a 10-15% wall plug efficiency. Thus that means an output of around 6mW. Then any optical element will reduce power by around 50% for each element. Assuming at most 2 optical elements (nu-naturally low number), that means the output can at max be 1.5mW.

In the answer you quoted, use this paragraph as a point of comparison:

For the sake of comparison, sunlight is one kilowatt per square meter and perhaps 5% of that is near infrared i.e. 700 to 1000 nanometers. Just going outside will expose you to much greater power densities of SWIR than the Kinect.

Also, remember that even though the generator is 60 mW (yes, I used the correct units), there is a series of diffusers, optics, and such so that at the very extreme of the exit aperture, the power density is <25 μW (again, note the symbol). The series of steps required to get at the 60 mW generator would indicate a willful intent to cause self harm, and be beyond simple mechanical failure.

Your initial assumption is incorrect.

My approach is as follows: - Assume 60mW output power is correct - Diffuser efficiency is 50% and therefore 50% of the energy is lost

The diffusers and optics reduce the power to <25 μW at the aperture. Run your math with that figure and you'll have an accurate representation.

The leakage current is the lowest current that you can (notionally) get away with designing a divider to handle. You can and should take many times that as "load current". This will then swamp leakage current effects.

Ginormous resistor values will lead to long rise and fall time constants (when combined with the IC's input capacitance and stray capacitances) which you do not want.

If you want mono-directional driving and level reduction only then a resistor divider will work fine.

"All" that is required is to ensure that

• Vout_high_max when divided down does not exceed Vin_high_max of the driven IC. (Better, input high and low worst case limits produce output voltages also in the required ranges.)

• Overall time constants are small when compared to your signal switching speeds OR

  • If you can't make them small they must be acceptable. eg when writing data mono-directionally a significant delay may acceptably occur in delay and data lines as long as it is consistent and about equal in all relevant signal leads. Problems may occur when handling bidirectionally as eg a delayed outgoing clock may causes data to be returned long after the clock has sampled the data "successfully".

1 Megohm x 1 picoFarad gives a time constant of 1 uS.
You can scale from there as required.

eg if you had 20 pF of stray and input capacitance and were using a 100 kOhm divider resistance (which would be high) then time constant 20 pF x 0.1 Megohm =~~~2 uS.

Referring to your 23A256 RAM as an example only

Input capacitance = 7 pF.
Max clock rise and fall time = 2 uS
BUT max clock rate can be 20 MHz (difficult) or 50 Ns total cycle time.

If all up capacitance is taken as 10 pF (7 pF input + 3 stray) then
- 100k drive resistance gives you 1 uS time constant on the driven line(!),
- 10 k drive resistance gives 100 Ns and
- 1k gives 10 Ns.

To achieve a 1k equivalent resistance with a 3.3V to 5 divider you can have Rupper = 1.4k and Rlower = 2.7k. Parallel resistance. = 0.92K and series res = 4.1 K so current at 5V ~= 1.2 mA.

The 10 nS time constant with resistors i the 2k range translates to rise and fall times of several tine constants - so these approach the times involved per half cycle at maximum speed. As these delays will be included in all down converted lines the differences in delays should be much smaller. If using bidirectional transfer this may cause a problem with eg data setup or hold times. Looking at the data sheet in each specific case is essential.

ie I'd expect a simple low value resistive divider to work in many mono-directional cases but care will be needed at higher clock speeds. Referring to diagram above.

• Reffective for tim contant calculations = R1//R2

• RseriesDC drive mA chck is R1 + R2.

• Vout = Vin x R2 / (r1+r2)

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